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Let $A$ and $B$ be two positive semidefinite $n \times n$ matrices. Does the following quadratic matrix equation have a solution in the set of real symmetric matrices?

$$XAX=B$$

It's a special case of the Riccati equation. I want just to prove the existence of such real matrix $X$.

Rodrigo de Azevedo
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Jef
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1 Answers1

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Let's assume $A$ is positive definite. Multiplying on left and right by $A^{1/2}$ (the positive definite square root of $A$) the equation becomes $$ (A^{1/2} X A^{1/2})^2 = A^{1/2} B A^{1/2}$$ Now $A^{1/2} B A^{1/2}$ is positive semidefinite, so has a positive semidefinite square root $C$, and we can take $X = A^{-1/2} C A^{-1/2}$ so that $C = A^{1/2} X A^{1/2}$.

Robert Israel
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  • well i think you reasoning is correct when A is positive definite but when its not if we take $X=A^{-1/2}(A^{1/2}BA^{1/2})^{1/2}A^{-1/2}$ we get that $XAX=P_ABP_A$ where $P_A$ is the projector of Range(A) on $\mathbb{R}^n$ – Jef Aug 18 '19 at 22:31
  • If $A$ is not positive definite, what do you mean by $A^{-1/2}$? Perhaps a pseudo-inverse? – Robert Israel Aug 19 '19 at 00:44
  • yeah the pseudo inverse in the sense of moore-penrosae – Jef Aug 19 '19 at 01:51